STRUCTURE OF Sn-112, Sn-114, Sn-115, Sn-116, Sn-117, Sn-118, Sn-119, Sn-120, Sn-122, Sn-124
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of electromagnetic laws in favor of wrong theories which could not lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for discovering the nuclear force and structure by applying the well-established laws of electromagnetism. (See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). Nuclear structure of Tin (Sn) with 28 blank positions Tin (Sn) is the element with the greatest number of stable isotopes (ten) (three of them are potentially radioactive but have not been observed to decay), which is probably related to the fact that 50 is a "magic number" of protons. 29 additional unstable isotopes are known, including the "doubly magic" tin-100 (100Sn) (discovered in 1994) and tin-132 (132Sn). Comparing the Sn-110 of 50 protons and 50 neutrons (even number ) with Cd of 48 protons (even number) one sees that the structure of Sn-100 is also of high symmetry. ( See my STRUURE OF Cd-106...Cd116 ). Here the additional p49n49 and p50n50 make the two symmetrical vertical rectangles with the n15p17 and the p16n18 which lead to the structure of 10 stable nuclides having extra neutrons less than N = 28. For example the lightest stable nuclide like the Sn-112 with S =0 is based on Sn-100 with S =0 and has 12 extra neutrons of opposite spins, while the heaviest one like the Sn-124 with S =0 has 24 extra neutrons. In the following diagram of Sn-100 with S =0 you can see the first alpha particle with p45, n45, p47 and n47 and the symmetrical one with the p46, n46, p48 and n48. You can see also the p45, n47, n46 and p48 by using the top view of the third horizontal plane. At the same top view of the third plane one can see also the 2 blank positions formed by p49 able to receive 2(n). Note that the p49 forms also a third blank position able to receive 1n near the p13 of the second horizontal plane. In the same way the p50 forms two blank positions for receiving 2(n) which are behind the p42 and p18 respectively. Also the p50 forms a third blank position able to receive 1n which is behind the p20 of the fifth horizontal plane. So in the structure of Sn-100 the two alpha particles contribute to the creation of 8 blank positions able to receive 4(n) + 4n. Especially 4(n) give the combinations of p45 and p41, of p47 and p42, of p46 and p43, and of p48 and p44. Whereas the combinations of p45 and p23, of p47 and p29, of p46 and p30, and of p48 and p24 give the extra 4n which are shown in the diagram of Sn near p23, p29, p30 and p24 respectively. Like the structure of Cd with two squares also in this new arrangement the p37n37 and the p39n39 as well as the p38n38 and the n40p40 give a total S = -2 +2 =0 Also they give 8 blank positions for receiving 8n of opposite spins. In other words in the structure of Sn-100 with S =0 in which the additional p49n49 and p50n50 give 4(n) and 2n the number N of blank positions is given by The two squares with 8n The first and sixth plane with 4(n) The second and fifth plane with 4{n} +6n The two alpha particles with 6(n) That is N = 8n + 4(n) + 4{n} + 6n + 6(n) of opposite spins Or N = 4{n} + 14n + 10(n) = 28 extra neutrons of opposite spins STRUCTURE OF Sn-112, Sn-114, Sn-116, Sn-118, Sn-120, Sn-122, Sn-124 WITH S =0 Since the two symmetrical alpha particles along with the additional p49n49 and p50n50 in the structure of Sn-100 give S=0 one concludes that the spin S = 0 of the above stable nuclides is due to the extra neutrons of opposite spins. Here the Sn-112 of 12 extra neutrons has 4{n} +8n of opposite spins, while the Sn-114 of 14 extra neutrons has 4{n} + 10n. In the same way we can find the extra neutrons of the next nuclides. For example the heaviest stable nuclide like the Sn-124 of 24 extra nucleons has 4{n} + 14n + 6(n) of opposite spins STRUCTURE OF Sn-115, Sn-117, AND Sn-119 WITH S = +1/2 These nuclides have an odd number of extra neutrons with the total spin S = +1/2. Therefore in order to reveal the structure of Sn-115 with 15 extra neutrons we conclude that it has 8 extra neutrons of positive spins and 7 extra neutrons of negative spins. Especially the Sn-115 has 2{n} +6n of positive spins and 2{n] + 5n of negative spins. Similarly the Sn-117 of 17 extra neutrons has 2{n] +7n of positive spins and 2{n} +6n of negative spins. Finally the Sn-119 of 19 extra neutrons has 2{n} + 7n +1(n) of positive spins and 2{n} +7n of negative spins. ' ' DIAGRAM OF Sn-100 WITH S = 0 FORMING 28 BLANK POSITIONS Here you see the p47n47 along with the p48n48, which make two symmetrical alpha particles of opposite spins . But you cannot see the additional p49n49 and p50n50 which makes the two symmetrical vertical rectangles with the n15p17 and the p16n18 respectively. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. Also 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Moreover the 4 extra neutrons (n) of the first and the sixth plane are not shown, while the extra neutrons 4n existing over the p31 and p32, and under the p21 and p22 along with the extra 4n existing near the p23, p24, p29 and p30, are shown. ' n40.........p40......n ' ' n........p38..........n38 H. square with n ' ' n31………p12.........n12.......p32' ' p31........n11.........p11…… n32 Sixth H. plane' ' n........ p29.........n10.........p10…… n30' ' n29………..p9..........n9 …….p30..........n Fifth H. plane' ' p47.......n27.........p8..........n8.........p28......... n48' ' n45.......p27.........n7..........p7........n28..........p46 Fourth H. plane' ' n47......p25.........n6.........p6..........n26...........p48' ' p45......n25……….p5..........n5……….p26.........n46 Third H. plane' ' n23………p4........n4………….p24..........n' ' n........p23……..n3………p3………..n24 Second H.plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First H. plane' ' n..........p37.......n37 ' ' n39.......p39.........n ' H. Square with 2n TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' ' ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22 ' ' n33.......p33..... (n)' ' ' TOP VIEW OF THE SECOND HORIZONTAL PLANE Here the {n} near the p14 fills the blank position formed by p14, p24 and p44. The 2n near p24 and p23 fill the blank positions formed by p24 and p48 as well as by p23 and p45. Moreover the blank position of {n} near p23 is formed by p23, p13 and p41. Finally the blank position of n near the p13 is formed byp13 and p49. That is we have 2{n} +4n and the same situation occurs at the fifth horizontal plane. ' ' ' n14.......p14........{n}' ' n23.......p4........n4.........p24......n' ' n.......p23........n3........p3.........n24' ' {n}.......p13......n13' ' n ' TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS HERE YOU SEE THE p41, n43, n42 AND p44 WHICH MAKE THE SQUARE OF THE CENTRAL PARALLELEPIPED. YOU SEE ALSO THE p45 AND n47 ALONG WITH THE n46 AND p48 WHICH MAKE THE SYMMETRICAL SQUARES OF THE TWO ALPHA PARTICLES. AT THE SAME PLANE ARE SHOWN ALSO THE DEUTERONS LIKE n16p16 AND p15n15 ALONG WITH THE ADDITIONAL p49 WITH 2(n) ' n42........p16......n16......p44' ' n47........p25........n6........p6........n26.........p48' ' p45........n25........p5........n5........p26........ n48' ' p41.......n15.......p15.......n43' ' (n)........p49.......(n)' ' ' Category:Fundamental physics concepts